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给定一个字符串s，找到其中最长的回文子序列。可以假设s的最大长度为1000。


示例 1:输入:    “bbbab”输出:    4一个可能的最长回文,"> 
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        <h1 class="title">leetcode516. 最长回文子序列-附动态规划状态图</h1>
        <div class="stuff">
            <span>四月 11, 2020</span>
            

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            <h1 id="leetcode516-最长回文子序列"><a href="#leetcode516-最长回文子序列" class="headerlink" title="leetcode516. 最长回文子序列"></a>leetcode516. 最长回文子序列</h1><hr>
<blockquote>
<p>给定一个字符串s，找到其中最长的回文子序列。可以假设s的最大长度为1000。</p>
</blockquote>
<blockquote>
<p>示例 1:<br>输入:<br>    “bbbab”<br>输出:<br>    4<br>一个可能的最长回文子序列为 “bbbb”。</p>
</blockquote>
<blockquote>
<p>示例 2:<br>输入:<br>    “cbbd”<br>输出:<br>    2</p>
</blockquote>
<p>本题和</p>
<p><a href="https://leetcode-cn.com/problems/palindromic-substrings//" target="_blank" rel="noopener">回文子串</a><br><a href="https://leetcode-cn.com/problems/longest-palindromic-substring//" target="_blank" rel="noopener">最长回文子串</a><br>思路类似。<br>用 <strong>dp[i][j]</strong> 表示 <strong>i~j</strong> 内回文子序列的最长长度。</p>
<h3 id="找出当前的状态怎么处理"><a href="#找出当前的状态怎么处理" class="headerlink" title="找出当前的状态怎么处理"></a>找出当前的状态怎么处理</h3><ul>
<li>当 <strong>s[i] == s[j]</strong> 时，只需用 <strong>dp[i+1][j-1]</strong> 加上2就是当前的最大长度，因为 <strong>dp[i+1][j-1]</strong> 已经含有回文子序列（上面定义），所以再加上2就行。(如果 <strong>i+1 &lt; j-1</strong> ，也就是子序列长度为2，需要特殊处理)</li>
<li>如果 <strong>s[i] != s[j]</strong> 。需要找出 <strong>i~j</strong> 里面最大的。而如果子序列最长，也就是 <strong>i-j</strong> 越大，他们的dp也会越大，因为字母多。所以只需要取 <strong>i-j-1</strong> 长度的序列就行。在 <strong>i~j</strong> 序列中，这样的序列有两个 <strong>i+1~j</strong> 和 <strong>i~j-1</strong> ，取这两个当中最大的。 <strong>dp[i][j] = Math.max(dp[i + 1][j],dp[i][j - 1]);</strong><h3 id="填表"><a href="#填表" class="headerlink" title="填表"></a>填表</h3>画出表格，看看应该怎么填表。<br><img src="https://img-blog.csdnimg.cn/20200410163350421.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzE1NzY0NDc3,size_16,color_FFFFFF,t_70#pic_center" alt="在这里插入图片描述"><br>根据式子dp[i][j] = Math.max(dp[i + 1][j],dp[i][j - 1]);，在填(1,3)时候会用到他的左面和下面。<br><img src="https://img-blog.csdnimg.cn/20200410163656398.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzE1NzY0NDc3,size_16,color_FFFFFF,t_70#pic_center" alt="在这里插入图片描述"><br>所以要先填不能用到左面和下面的点，也就是说当前填的点必须用到填过的，不能用到没填过的。<br>填表的方式应该很多，我选择的是从最后一行开始向右遍历，再逐步向上遍历。<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">longestPalindromeSubseq</span><span class="params">(String s)</span> </span>&#123;</span><br><span class="line">      <span class="keyword">int</span> n = s.length();</span><br><span class="line">      <span class="keyword">int</span> max = <span class="number">1</span>;</span><br><span class="line">      <span class="keyword">int</span>[][] dp = <span class="keyword">new</span> <span class="keyword">int</span>[n][n];</span><br><span class="line">      <span class="keyword">for</span> (<span class="keyword">int</span>  i = n - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i--) &#123;</span><br><span class="line">          dp[i][i] = <span class="number">1</span>;</span><br><span class="line">          <span class="keyword">for</span> (<span class="keyword">int</span> j = i + <span class="number">1</span>; j &lt; n; j++) &#123;</span><br><span class="line">              <span class="keyword">if</span> (s.charAt(i) == s.charAt(j)) &#123;</span><br><span class="line">                  <span class="keyword">if</span> (i + j == <span class="number">1</span>) &#123;</span><br><span class="line">                      dp[i][j] = <span class="number">2</span>;</span><br><span class="line">                  &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                      dp[i][j] = dp[i + <span class="number">1</span>][j - <span class="number">1</span>] + <span class="number">2</span>;</span><br><span class="line">                  &#125;</span><br><span class="line">              &#125; <span class="keyword">else</span>&#123;</span><br><span class="line">                  dp[i][j] = Math.max(dp[i + <span class="number">1</span>][j],dp[i][j - <span class="number">1</span>]);</span><br><span class="line">              &#125;</span><br><span class="line">              </span><br><span class="line">              max = Math.max(max, dp[i][j]);</span><br><span class="line">          &#125;</span><br><span class="line">      &#125;</span><br><span class="line">      </span><br><span class="line">      <span class="keyword">return</span> max;</span><br><span class="line">  &#125;</span><br></pre></td></tr></table></figure>
<strong>leetcode 77/100</strong></li>
</ul>

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